Question

If $x=\frac{2t}{1+t^2},y=\frac{1-t^2}{1+t^2}$, then $\frac{dy}{dx}$ equals:

• A. $\frac{2t}{t^2+1}$
• B. $\frac{2t}{t^2-1}$
• C. $\frac{2t}{1-t^2}$
• D. None of these

Answer 1

The correct option is B

$\frac{2t}{t^2-1}$

Explanation for the correct option:

Step 1: Find $\frac{dy}{dt}$

$\begin{array}{rcl}y& =& \frac{1-t^2}{1+t^2}\\ \Rightarrow y& =& \frac{2-\left(1+t^2\right)}{1+t^2}\\ \Rightarrow y& =& \frac{2}{1+t^2}-1\\ & \Rightarrow & \frac{dy}{dt}=-\frac{2}{{\left(1+t^2\right)}^2}\times 2t\\ & \Rightarrow & \frac{dy}{dt}=-\frac{4t}{{\left(1+t^2\right)}^2}\end{array}$

Step 2: Find $\frac{dx}{dt}$

$$\begin{gathered} x=\frac{2t}{1+t^2} \\ \Rightarrow \frac{dx}{dt}=\frac{\left(1+t^2\right)2-2t\left(2t\right)}{{\left(1+t^2\right)}^2} \\ \Rightarrow \frac{dx}{dt}=\frac{\left(1+t^2\right)2-2t\left(2t\right)}{{\left(1+t^2\right)}^2} \\ \Rightarrow \frac{dx}{dt}=\frac{2\left(1-t^2\right)}{{\left(1+t^2\right)}^2} \end{gathered}$$

Step 3: Find $\frac{dy}{dx}$

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{{\displaystyle \frac{-4t}{{\left(1+t^2\right)}^2}}}{{\displaystyle \frac{2\left(1-t^2\right)}{{\left(1+t^2\right)}^2}}}\\ & =& \frac{-2t}{1-t^2}\\ & =& \frac{2t}{t^2-1}\end{array}$

Hence, option B is correct.